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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## If the farmer sells 75 of his chickens, his stock of feed wi ##### This topic has 3 expert replies and 1 member reply ### Top Member ## If the farmer sells 75 of his chickens, his stock of feed wi ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have? A. 60 B. 120 C. 240 D. 275 E. 300 OA E Source: Manhattan Prep ### Top Member Legendary Member Joined 02 Mar 2018 Posted: 1104 messages Followed by: 2 members Top Reply $$Let\ no\ of\ chicken=x$$ $$Let\ no\ of\ days=d$$ If the farmer sells 75 of his chicken, his stock of feed will last for 20 more days than planned If he buys 100 more chicken, he will run out of feeds 15 days earlier than planned $$Amount\ of\ feed\ =xd$$ $$xd=\left(x+100\right)\left(d-15\right)$$ $$where,\ xd=\left(x-75\right)\left(d+20\right)\$$ $$\left(x-75\right)\left(d+20\right)\ =\left(x+100\right)\left(d-15\right)$$ $$\frac{\left(d+20\right)}{\left(d-15\right)}=\frac{\left(x+100\right)}{\left(x-75\right)}$$ $$\left(x=5d\right)$$ $$xd=\left(x-75\right)\left(d+20\right)$$ $$5d^2=\left(5d-75\right)\left(d+20\right)$$ $$d^2=\left(d-15\right)\left(d+20\right)$$ $$d=60$$ $$x=5d$$ $$x=5\cdot60=300$$ $$answer\ is\ OptionE$$ ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2791 messages Followed by: 18 members Upvotes: 43 Top Reply BTGmoderatorDC wrote: If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have? A. 60 B. 120 C. 240 D. 275 E. 300 OA E Source: Manhattan Prep Let c = the number of chickens the farmer currently has and let d = the number of days he can feed the chickens. Notice that the number of chickens and the number of days the chicken can be fed with the constant amount of food the farmer has are inversely proportional. We can create the equations: cd = (c - 75)(d + 20) and cd = (c + 100)(d - 15) Simplifying the equations, we have: cd = cd - 75d + 20c - 1500 cd = cd + 100d - 15c - 1500 Now adding the two equations, we have: 0 = -175d + 35c 175d = 35c d = 35c/175 = c/5 Recall that cd = cd - 75d + 20c - 1500 or, simply, 0 = -75d + 20c - 1500. Now substitute c/5 for d, obtaining: 0 = -75(c/5) + 20c - 1500 0 = -15c + 20c - 1500 1500 = 5c 300 = c Answer: E _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 BTGmoderatorDC wrote: If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have? A. 60 B. 120 C. 240 D. 275 E. 300 Source: Manhattan Prep $$\left( {\# {\rm{chicken}}} \right)\left( {\# days} \right) = {\rm{constant}}$$ $$? = x = \# {\rm{chicken}}$$ $$\left\{ \matrix{ \left( {x - 75} \right)\left( {d + 20} \right) = \left( {x + 100} \right)\left( {d - 15} \right)\,\,\,\,\left( {\rm{I}} \right) \hfill \cr \left( {x - 75} \right)\left( {d + 20} \right) = xd\,\,\,\,\left( {{\rm{II}}} \right) \hfill \cr} \right.$$ $$\left( {\rm{I}} \right)\,\,\, \Rightarrow \,\,\,xd + 20x - 75d - 75 \cdot 20 = xd - 15x + 100d - 15 \cdot 100$$ $$\Rightarrow \,\,\,35x - 175d = 20\left( {75 - 5 \cdot 15} \right) = 0\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,35} \,\,\,\,\,x = 5d\,\,\,\left( {{\rm{III}}} \right)$$ $$\left( {{\rm{II}}} \right)\,\,\, \Rightarrow \,\,\,xd + 20x - 75d - 75 \cdot 20 = xd\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {{\rm{III}}} \right)} \,\,\,\,\,20x - {{75 \cdot x} \over 5} = 75 \cdot 20$$ $$\Rightarrow \,\,\,5x = 75 \cdot 20\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 15 \cdot 20 = 300\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {\rm{E}} \right)$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15348 messages Followed by: 1864 members Upvotes: 13060 GMAT Score: 790 BTGmoderatorDC wrote: If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. How many chickens does the farmer have? A. 60 B. 120 C. 240 D. 275 E. 300 Total amount of feed = (number of chickens)(number of days). The number of chickens is INVERSELY PROPORTIONAL to the number of days. If the number of chickens DOUBLES, then the feed will last for 1/2 the number of days. If the number of chickens TRIPLES, then the feed will last for 1/3 the number of days. Let f = feed, c = chickens, and d = days. Thus: f = cd. Quote: If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. The colored portions and the answer choices indicate that the problem is constrained to POSITIVE INTEGERS. The portions in blue imply that the value of c -- a value that must divide evenly into f -- is probably a multiple of 75 and 100. The LCM of 75 and 100 is 300. Thus, it is almost certain that c = a multiple of 300. The portions in red imply that the value of d -- a value that also must divide evenly into f -- is probably a multiple of 20 and 15. The LCM of 20 and 15 is 60. Thus, it is almost certain that d = multiple of 60. Of the five answer choices, only E is a multiple of 300. Test E, letting c = 300 and d = 60, implying that f = 300*60 = 18,000. When the correct answer is plugged in, the value of f will remain constant. E: 300 If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned. (c-75)(d+20) = (300-75)(60+20) = (225)(80)= 18,000. if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. (c+100)(d-15) = (300+100)(60-15) = (400)(45) = 18,000. Success! In each case, the value of f remains constant at 18,000. The correct answer is E. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! 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